Question

An electron is accelerated by a potential difference of 10 kV. What is the de Broglie wavelength associated?

a. Shorter than 1 nm
b. Between 1 nm and 10 nm
c. Between 10 nm and 100 nm
d. Longer than 100 nm

Answer 1

The correct answer (a) Shorter than 1 nm. The de Broglie wavelength of an electron accelerated by a potential difference of 10 kV is shorter than 1 nm since the calculated wavelength is approximately 0.0123 nm.

Step-by-step explanation:

To calculate the de Broglie wavelength of an electron that has been accelerated by a potential difference, we use the equation λ = h / p, where λ is the wavelength, h is Planck's constant (6.626 x 10-34 J·s), and p is the momentum of the electron.

Since the electron is accelerated from rest, its kinetic energy (KE) is equal to the energy acquired from the electric potential, which is KE = eV, where e is the charge of the electron (1.6 x 10-19 C) and V is the potential difference.

Thus, the momentum p can be calculated using the equation p = √(2mKE), where m is the mass of the electron (9.11 x 10-31 kg).

Calculating the momentum for our question:

p = √(2m eV)

= √(2 × 9.11 x 10-31 kg × 1.6 x 10-19 C × 10 x 103 V)

= √(2.9056 x 10-24 J)

= 5.39 x 10-13 kg m/s.

Now, we calculate the wavelength

λ = h / p

= (6.626 x 10-34 J·s) / (5.39 x 10-13 kg m/s)

= 1.23 x 10-11 m or 0.0123 nm.

Therefore, the de Broglie wavelength of the electron accelerated by a potential difference of 10 kV is shorter than 1 nm, making the correct answer (a) Shorter than 1 nm.